fulltest iv advanced paper 2 answer sol aits jeea ft iv paper 2

8/11/2019 Fulltest IV Advanced Paper 2 Answer Sol Aits Jeea Ft IV Paper 2

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AITS-FT-IV-(Paper-2)-PCM(S)-JEE(Advanced)/13

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1

ANSWERS, HINTS & SOLUTIONS

FULL TEST IV(Paper-2)

Q. No. PHYSICS CHEMISTRY MATHEMATICS

1. C C A

2. A B C

3. D B A

4. B C C

5. A D C

6. C C B

7. A A B

8. A C D

9. D C C

10. C A B

11. A C B

12. C B B

13. A B D

14. B D B

15. A, B, C A,C A, B

16. A, B, C A,C A, B, C, D

17. A, D B,D A, B, C

18. A, B, D B,D C, D

19. A, C, D A, B, D A, C

20. C, D B, C, D A, C

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2

PPhhyyssiiccss PART ISECTION A

1. A is at rest instantaneously (for no slipping at A)

VB VA=

1

2 B

0V

R

(2R)2

VC VA=1

2B 0

V

R

( 3 R)2

VB VC=1

2B 0

V

R

[R2] = 0BV R

2

2. force = (P P0) A = Ma

by P1V1= P2V2

P0hA = P(h x) A

P = 0P h

h x

So force = 0 0P h

Ph x

A

M = a

a = 0P Ax

M(h x) 0

P Ax

Mh; (for small x)

T = 20

Mh

P A

3. restoring= (mg sinL + (mg sinL

2

+

L L KL L

K sin sin4 4 2 2

+

restoring=23mg 5KL

2 16

+

(: small)

frequency of oscillation) =

23mg 5kL1 2 16

2

+

where = mL2+2 2mL 4mL

3 3=

4. From law of radioactive decay.

Activity Number of nuclear of that nucleus present in the sample.Activity does not depend upon the rate of the formation of nucleus.In given reaction B decays into E and C.Total activity of B = rate of decay of B into E + rate of decay of B into C = l5N2+ l3N2

5. The EMF will be induced in the loop, only when there is a change in the flux through the loop.When the loop was completely inside the field region, for some time (say t1) there is no change in

the flux, hence that time would not be included in the calculation.


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3

The time for which EMF was induced =b b 2b

v v v+ =

6. That wire will break first whose stress reaches the breaking stress (for thatmaterial) first.

Stress =2

F F

A r=

where F is the tension in the wire. F is same in both the wires.

Thus stress 2

1

rfor the wire of smaller radius, the stress is more.

Thus the wire with smaller radius will break first.If rA< rB, then wire A will break first.

If rA> rB, then wire B will break first.

If rA= rB, either of them could break.

8. As the magnetic field is uniform and the particle is projected in a direction perpendicular to the

field, it will describe a circular path. The particle will not hit the yz-plane if the radius of the circle is

smaller than d. For the maximum value of v, the radius is just equal to d. Thus =mv

dBq

.

=Bqd

vm

9. F = A 0u u

y

u =F

Ay + u0

u =2

1

10 1 y + 1 = 100

y + 1

10. The velocity of upper plate is same as that of the layer in contact with the plate i.e. layer at y = 2cm.u = 100 (0.02) + 1 = 3 m/s

11. T2r3 (Kepler's law)


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4

2 3

A A

2 3B B

T r

T r= =

3

3

r 1

64(4r)= Get A

B

T 1

T 8=

12. For geostationary satellite, time period = 1 planet day (by def.)Let T = 1 planet dayT0= 1 planet year

Now T2 =2

3G

4r

Gm

=

234 mr

Gm M

=2

3 30

4r T

GM

= or T = T0

13-14. =k

M, =

k

M m+=

2500

25= 10 rad/sec

= M

M m+

x1= A sin 1

Potential energy stored in the spring at (t = t1) =2 2 2

1 1

1 1kx M sin

2 2=

Let v and vbe the velocity of system just before collision and just after collision, so using COLM

v=( )

Mv

M m+=

( )

MA cos

M m

+

Total energy after collision = PE + KE = ( )2 21

1 1kx M m v

2 2+ +

=2

2 21 M msinMA2 M m

+ +

=2

21 M msinkA2 M m

+ +

= 8 J

( )

2 2 2 21 1

M m A MA2 2 + =

2M msin

M m

+

+

15.2

max 1 2

2min 1 2

I (a a )

I (a a )

+=

1 1

2 2

a I 4 2

a I 9 3= = =

max

min

I25

I =

If an identical paper is pasted across second slit shifted C.B.F. will back in central point.

( )tD

shift 1

d

=

fringe widthn D

d

=

shift ( 1)tn 15

fringewidth

= = =

.

16. The change in magnetic flux is zero, hence the current in the ring will be zero.

17. Both (A) and (D) depend on g.


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5

18.n(n 1)

6 n 42

= =

If the initial state were n = 3, in the emission spectrum, no wavelengths

shorter than 0would have occurred. This is possible if initial state were n = 2 (A), (B) and (D)

19. From COE (A) is correct.

The force outside the earth varies as inverse square of the distance.

motion is not simple harmonic. However, from symmetry of motion, the motion will beperiodic.

(C) is correct.From COE

21 GMm 3GMmmv2 2R 2R

=

2GMv

R =

(A), (C) and (D) are correct.

20. The current through R1is constant.


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6

CChheemmiissttrryy PART II

SECTION A

1. pH at isoelectric point = 1 2pKa pKa 4 9 6.52 2+ += =

2. Net reaction does not involve any aqueous species.

5. ( ) ( )2

32 4Cr SCN NH

+ , shows geometrical (or cis-trans) and linkage isomerism.

6.

CH3 C CH2 C O CH2 CH3

O O

(i) EtO

(ii)

H C H

O CH3 C CH C O CH2 CH3

O O

CH2 OH

2, OH CH3 C

O

CH C O CH2

O CH3

CH3 C C C O CH2 CH3

O O

CH2

( )Michael

CH3 C CH C O CH2 CH3

O O

CH2

CHCH3 C C O CH2 CH3

O O

EtO

OH

O

C

O

O CH2 CH3

C

O

O CH2 CH3

Li /+

CH3

O

C

O

O Et

2 2H O CO Et OH+ + +

7.4 2 3 3 2FeSO Fe O SO SO

+ +

8. Energy of e in H-atom = - E

Energy of e in Li+2

= - 9E

Energy supplied by photon = 21

IE mv2

+

IE E=

2p

1E 9E mv2 = +

( )p2 E 9Ev

m

=

9. In reverse osmosis water moves from more concentrated to less concentrated solution

10. = iCRT, (i=2)


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7

11. Let conc. of both reactants at equilibrium is x M2

2

2

PbEDTA HKeq

x

+ =

x = 103

12. Chelation increases stability of complexes

14.

2 3 2aq.K CO H /H OHCN

heatP Q R S

+

+

O

OH

O

2H O

O

OH

O

OH

OHOH

O2H /H O

heat

+

CN

OH

OH

( )S

HCN

HOH

O

( )R

2 3aq. K CO

O

H H

+H

O

+

( )P ( )Q

15.

Cl

Cl

KCN Cl

CN

2H PtCl

NH2

16. Trans-2-butene 3 3 3 4CF CO H H O HIO

3epoxide diol 2CH CHO+

17. He and H2are not hydrogen like particles

18. Radius of cation is smaller than parent atom and that of anion is more than parent atom

19. (A)

H2C

CH O

O

I

OC

O

CH2 I

I2will attack the double bond to form a cyclic halonium ion. It then undergoes on intramolecularreaction. Formation of five membered ring is favoured because (i) attack by a nucleophile on ahalonium ion normally takes place at the more substituted site.(ii) Formation of five membered ring is more probable.


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8

(B)

O

C

O

CH2Br

O

O

Br+

CH3

(C)

Option(C)givesO

C

C

O

O

(D)Option(D)gives C O

1, 4 and 1,5-dicarboxylic acids give anhydrides whereas 1,6 and 1, 7-diacids give ketones when

heated with P2O5.


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9

MMaatthheemmaattiiccss PART IIISECTION A

1. Normal to y2= 4ax with slope m is

y = mx 2am am3 ..... (1)

(1) touches x2

y2

= a2

if(2am am3)2= a2(m21)i. e. if m

6+ 4m

4+ 3m

2+ 1 = 0 ..... (2)

(2) is a sixth degree equation in m with no odd power terms and whose coefficients are allpositive.

(2) has no real root|S| = 0

2. Given x + y + z = 5 (1)and x

2+ y

2+ z

2= 9 (2)

We have yz = ( ) ( )2 2 21 y z y z2

+ +

= ( ) ( )2

21

5 x 9 x2 (using) (1) & (2))

= x25x + 8

y, z must be the roots of the equation

( ) ( ) ( )2 2u 5 x u x 5x 8 0 3 + + = As y, z are real, we must have discriminant of (3) 0(5 x)2 4 (x2 5x + 8) 03x2 10x + 7 0(3x 7) (x 1) 0

x 7

1,3

Number of integer values that x can take is 2

3. 2 sin2x + sin

22x = 2

2(1 sin2 x) + 4 sin2x cos2 x = 0cos2x.cos 2x = 0 (1)sin 2x + cos 2x = tan x

2 sin x cos2 x + cos 2x cos x = sin xsin x cos 2x + cos 2x cos x = 0cos 2x (sin x + cos x) = 0 (2)Common roots of (1) and (2) are given by cos 2x = 0

( )2n 1

x ,n N4

=

In [0, 4] we have x =

n

,4

where n = 1, 3, 5,,15 (3)Now, 2 sin2x + sin x 02 sin2x sin x 0(sin x) (2 sin x 1) 0

sin x [1, 0] or1

sinx 12

(4)

Among the 8 values given by (3), only 4 values, i.e.,5 7 13 15

x , , ,4 4 4 4

= satisfy (4)


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10

4. We have1

w zz

= +

Take z = ei( |z| = 1).

We have i iw e e 2cos = + = We know that w is real and | w | 2

5. The scalar triple product of the three given vectors is

( ) ( )2 2 2

2 2 2

2 2 2

x 1 2 x 1 3 x 1

2x 1 2x 1 x

3x 2 x 4 x 1

+

+ + +

( )2 2 2 22 2 2

1 2 3

x 1 2x 1 2x 1 x

3x 2 x 4 x 1

= +

+ + +

( )2 2 2 22 2 2

1 2 0

x 1 2x 1 2x 1 5x

3x 2 x 4 5x 7

= +

+ + +

(using C3C3 + C2+ C1)

( )2 2 2 2

2 2 2

1 0 0

x 1 2x 1 2x 3 5x

3x 2 5x 5x 7= +

+ +(using C2C2 2C1)

= ( )( )2 4 2x 1 15x x 21 + + Vector are noncoplanar Scalar triple product 0 x2 1 0 (15x4+ x2+ 21 0) x 1

6. We have ( ) ( )1/ n

nf x a x= so that

( ) ( )1/ n

nf f x f a x =

( )1/ n

n1/ n

na a x =

( )( )1/ n

na a x=

( ) ( )2mtimes

g x f f .....f x x = =

( )( ) ( )( )( ) ( )1/ n 1/ n

n n n n

dx dx

x 1 xg x 1 g x

=++

( )

1/ nn 1 n

dx

x 1 x+ =

+ = 1/ n

dt

nt (using t = 1 + xn)

1/ n 11 tc

1n1

n

+

= + +

= ( )1

1n n

11 x c

1 n

+ +

7. Given ( ) ( )y f x ,z g x= = we have

( )

( )

f xdy dy / dx

dz dz / dx g x

= = and

2

2

d y d f dx

dx g dzdz

= ( )2

g f f g 1

. gg

=

8. ( )f x is continuous at x = 3

( )f 3 = ( ) ( )f 3 f 3+=

Now ( ) ( )22

2 2x 3 x 3

a x x 6a | x x 6 |f 3 lim lim

6 x x 6 x x

= =

+ + = a


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11

( )x 3 x 3

x [x] x 3f 3 lim lim 1

x 3 x 3+ ++

= = =

( )

x

3

x 3

x tdt0

f 3 b lim formx 3 0

=

x

2

3

x 3

x tdt

b lim1

+

=

= 9b

a = 9b = 1 b =1

9and a = 1

9. ( )n n

n r

r 1 r 1

V S 2 r n n 1

= =

= = = +

n

2n n

V n 1lim lim 1

nn

+ = =

10. (i) Vn= n (n + 1) V1< V2< V3< .....

Statement 2 is true

(ii)( )

n n

nr r 1r 1 r 1

1 1 1Q

S S 4 r r 1+= =

= =+

n

r 1

1 1 1 1 11

4 r r 1 4 n 1=

= = + +

Q1< Q2< Q3< .Statement 1 is false

11. We have

( )x

x x

2a , x 0f x

a a , x 0

( )x x

e ef x a log a a log a

= + = (ax

ax

) logea

2x

ex

a 1

log aa

= > 0 for x > 0

( )f x is monotonic increasing for x > 0

For x < 0, ( ) xf x 2a= ( ) x ef x 2a log a = < 0

( )f x is monotonic decreasing for x < 0

12. Required area ( )0 1

x x x

1 0

2a dx a a dx

= + + 0 1

x x x

e e1 0

2a a a

log a log a

+ = +

( )e e

1 1 12 2a a 0

log a log a a

= + +

2

e

3a 2a 1

alog a

=

15. The equation of the line is

1 1 2 2

3 3 3 3b y a x a b 0+ + =

This can be written as

12 1

33 3

1

3

ay x a b

b

=

12 1

33 3

1

3

am , c a b

b

= =

Consider the parabola y2= 4ax


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12

12 1

33 3

1 1

3 3

1

3

a a aba b c

ma a

b

= = = =

Condition for tangency for the parabola y2

= 4ax is satisfiedThe given line touches y2= 4axSimilarly the line touches x

2= 4by

16. ( )1

5x iy a ib+ = +

x + iy = ( a + ib)5 = a5+ 5a4(ib) + 10a3(b2) + 10a2(ib3) + 5ab4+ ib5Equating the real and imaginary parts x = a

5 10a

3b

2+ 5ab

4y = 5a4b 10a2b3+ b5

x yp

a b= = (a410a2b2+ 5b4) (5a4 10a2b2+ b4) = 4a4+ 4b4= 4 (a4b4)

= 4(a2 b2) (a2 + b2) = 4(a + b)( a b) ( a +ib) (a ib)a b, a + b, a + ib and a ib are all factors of p.

17. Since 2 2 20 ,0 sin , cos , tan 14< < < <

( ) 2k 22

k 0

1A sin sec

1 sin

=

= =

(B) 2k 22

k 0

1cos cosec

1 cos

=

= =

(C)2

2k

2 2 2k 0

1 costan

1 tan cos sin

=

= =

2cos

cos2

=

(D) cot > 1 in 0 < < .4 Hence, 2k

k 0

cot

=

does not exist.

19. y x x3 9 3 9 3 0= > x 2< Similarly, y < 2

20. Area ( )2 3c.c c

T2 2

= =

Area ( )c3

2

0

cR x dx

2=

3 3 3c c c

2 3 6= =

( )

( )

3

3c 0 R 0

Area T c 6lim lim . 3

Area R 2 c+ + = =

fulltest iv advanced paper 2 answer sol aits jeea ft iv paper 2

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